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Question

If the value of π/2π/2x21+tanx+1+tan2xdx is π3a. Then a=

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Solution

I=π/2π/2x21+tanx+1+tan2xdx
Using the property,
ba f(x) dx=ba f(a+bx) dx
I=π/2π/2x21tanx+1+tan2xdxOn adding these two integrals we get, 2I=π/2π/2x2(2+21+tan2x)(1+1+tan2x)2tan2xdx2I=π/2π/2x2dx=[x33]π/2π/22I=13×2π38I=π324
a=24

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